Have you tried replacing 2 by (1+lambda)? Alternatively, for large lambda and alpha far enough away from 1/2, some nice estimates occur by using geometric series approximations. Gerhard "Ask Me About System Design" Paseman, 2012.004.11
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Gerhard PasemanApr 11 '12 at 10:12

There are many estimates here. Is there context as to what kind of estimate you are looking for?
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Daniel ParryApr 11 '12 at 18:34

3 Answers
3

Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$.

If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the largest terms are approximately in geometric progression so you can get it quite accurately by computing the ratio.

If $α>λ/(λ+1)$, almost all of the complete binomial expansion is present, so the sum equals $(1+o(1))(1+λ)^n$.

If $α≈p$, then Russell's normal approximation will be good. (This needs some work to clarify whether the geometric approximation of the lower tail is good right up to the point where the normal approximation begins to be good. I think it is.)

The last assertion is actually not true, as stated: say, if $\lambda=1$ and $\alpha$ is just a tiny bit larger than $1/2$, then the sum is not equal to $(1+o(1))2^n$ (it is half of this quantity).
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SevaApr 11 '12 at 14:02

Perhaps you are not taking $\alpha$ to be constant as $n\to\infty$. Apart from the scaling factor of $2^n$, it is a binomial distribution $Bin(n,1/2)$, which has a standard deviation of $\sqrt{n}/2$. So if $\alpha=1/2+\epsilon$ for fixed $\epsilon>0$, the upper limit of your sum is $2\epsilon\sqrt{n}$ standard deviations above the mean, so it cuts off a negligible portion of the sum from 0 to $n$. To asymptotically get half of $2^n$ you need $\alpha=1/2+o(n^{-1/2})$, which is not constant except for $\alpha=1/2$.
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Brendan McKayApr 12 '12 at 2:08

What happens when $\alpha=c/n$ for $c$ a constant? Can one bound the sum in terms of $c$, $\lambda$ and $n$?
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BullmooseNov 22 '13 at 4:33

@Bullmoose: In that case the sum is strongly dominated by its largest term.
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Brendan McKayNov 22 '13 at 4:50

Your sum can also be thought of as the first $\alpha n$ terms in a binomial distribution with probability of success $p=1-\frac1{\lambda+1}$. So, it is closely approximated by a normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$, i.e., $$\sum_{k=0}^{\alpha n} \binom nk \lambda^k\approx (1-p)^{-n}\Phi\left((\alpha-p)\sqrt{\frac n{p(1-p)}}\right),$$
where $\Phi$ is the cumulative standard normal distribution.

Sums of this sort are estimated by their largest summand, and the resulting
estimate will depend on the relation between $\alpha$ and $\lambda$.

The ratio of the $(k+1)$th and the $k$th terms of the untruncated sum is
$$ \lambda \frac{n-k}{k+1}, $$
showing that the sequence of summands is unimodal, with the maximum value
attained for $k$ about $\frac{\lambda}{\lambda+1}\,n$. Consequently, if
$\alpha<\lambda/(\lambda+1)$, then your sum is between $\binom
n{\lfloor\alpha n\rfloor}\lambda^{\lfloor\alpha n\rfloor}$ and $n\binom n{\lfloor\alpha n\rfloor}\lambda^{\lfloor\alpha n\rfloor}$, which is
$2^{H((\alpha)+\alpha\log_2\lambda+o(1))n}$; similarly, if $\alpha>\lambda/(\lambda+1)$, then the
sum is $2^{(H(\lambda/(\lambda+1))+(\lambda/(\lambda+1))\log_2\lambda+o(1))n}=(\lambda+1)^{1+o(1)}$. (Both estimates assume that $\alpha$ and $\lambda$ are fixed, and $n\to\infty$.)